Question 306060
Solve for x:
{{{Log[5](2x-7) = 3}}} Rewrite this in exponential form:{{{Log[b](x) = y}}} converts to {{{b^y = x}}}, so...
{{{Log[5](2x-7) = 3}}}--->{{{5^3 = 2x-7}}} Substitute {{{5^3 = 125}}}
{{{125 = 2x-7}}} Add 7 to both sides.
{{{132 = 2x}}} Divide both sides by 2.
{{{highlight(x = 66)}}}